Results 1 to 6 of 6

Math Help - Pappus's Theorem and Some Integrals

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    28

    Pappus's Theorem and Some Integrals

    Hey guys, I was wondering if it would be possible for someone to check my work on a few calculus problems and, if they're wrong, to tell me what I'm doing wrong. Thank you very much in advance.

    Problem 1. R is the region bounded by y = sinx and the lines x = 0 and x = \frac{\pi}{2}.

    a. Calculate the center of mass of the region R.

    My Solution

    \overline{x} = \frac{\int^{\frac{\pi}{2}}_{0}xsinx}{\int^{\frac{\  pi}{2}}_{0}sinx} dx

    u = x, du = dx, dv = sinx dx, v = cosx

    -xcosx + \intcosx dx = -xcosx + sinx

    \frac{[-xcosx + sinx]^{\frac{\pi}{2}}_{0}}{[-cosx]^{\frac{\pi}{2}}_{0}} = \frac{1 - 0}{0 + 1} = 1

    \overline{y} = \frac{\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\sin^2(x)  }{\int^{\frac{\pi}{2}}_{0}sinx} dx = \frac{\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\frac{1 - cos(2x)}{2}}{[-cosx]^{\frac{\pi}{2}}_{0}} dx = \frac{\frac{1}{4}\int^{\frac{\pi}{2}}_{0}1 - cos(2x)}{0 + 1} dx = \frac{\frac{1}{4}[x - \frac{1}{2}sin(2x)]^{\frac{\pi}{2}}_{0}}{1} = \frac{\frac{1}{4}(\frac{\pi}{2}) - 0}{1} = \frac{\frac{\pi}{8}}{1} = \frac{\pi}{8}

    b. Use Pappus's Theorem to calculate the volume of the solid of revolution if R is revolved about the line x = 5.

    My Solution

    Area of region R = 1 (from problem 1a)

    V = 2\pi(5 - \frac{\pi}{2})(1) = 2\pi(\frac{10 - \pi}{2}) = 10\pi - \pi^2

    c. Calculate the volume of the solid if R is revolved about the line y = -4.

    My Solution

    Area of region R = 1 (from problem 1a)

    V =  2\pi(\frac{\pi}{8} + 4)(1) = 2\pi(\frac{\pi + 32}{8}) = \frac{\pi^2 + 32\pi}{4}

    Problem 2.
    Calculate each antiderivative.

    a. \int\frac{1}{\sqrt{1-x^2}} dx

    My Solution


    x = \sin\theta, dx = \cos\theta d \theta, \theta = \arcsin{x}

    = \int\frac{1}{\sqrt{1 - \sin^2(\theta)}}\cos(\theta) d \theta = \int\frac{1}{\sqrt{\cos^2(\theta)}}\cos(\theta) d \theta = \intd \theta = \theta + C = \arcsin{x} + C

    b. \int\frac{x}{\sqrt{1 - x^2}} dx

    My Solution

    u = 1 - x^2, du = -2x dx, dx = \frac{-1}{2x} du

    = \int\frac{x}{u^\frac{1}{2}}(\frac{-1}{2x}) du = \frac{-1}{2}\int \frac{1}{u^\frac{1}{2}} du = \frac{-1}{2}\int u^\frac{-1}{2} du = -\sqrt{u} + C = -\sqrt{1 - x^2} + C

    c. \int\frac{x^2}{\sqrt{1-x^2}} dx

    My Solution

    x = sin\theta, dx = cos\theta d\theta, \theta = \arcsin{x}

    = \int\frac{\sin^2(\theta)}{\sqrt{1 - \sin^2(\theta)}}cos\theta d\theta = \int\frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)}}co  s(\theta) d \theta = \int\frac{\sin^2(\theta)}{cos(\theta)}cos(\theta) d \theta = \int\sin^2(\theta) d\theta = \int\frac{1 + cos(2\theta)}{2} d\theta = \frac{1}{2}\int 1 + cos(2\theta) d\theta = \frac{1}{2}(\theta + \frac{1}{2}\sin(2\theta)) = \frac{1}{2}\arcsin{x} + \frac{1}{4}\sin(2x) + C

    d. \int\frac{x^3}{\sqrt{1 - x^2}} dx

    My Solution

    x = \sin(\theta), dx = \cos(\theta) d\theta, \theta = \arcsin{x}

    = \int\frac{\sin^3(\theta)}{\sqrt{1 - \sin^2(\theta)}}\cos(\theta) d\theta = \int\frac{\sin^3(\theta)}{\sqrt{\cos^2(\theta)}}\c  os(\theta) d\theta = \int\frac{\sin^3(\theta)}{cos(\theta)}\cos(\theta) d\theta = \int\sin^3(\theta) d\theta = \int\sin(\theta)(\frac{1 + cos(2\theta)}{2}) d\theta = \frac{1}{2}\int\sin(\theta)(1 + cos(2\theta)) d\theta = \frac{1}{2}\int\sin(\theta) + \sin(\theta)\cos(2\theta) d\theta = \frac{1}{4}[-\cos(\theta) + \cos^2(2\theta)] + C = \frac{1}{4}[\sqrt{1 - x^2} - 2(1 - x^2)] + C

    Once again, thank you very much in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by mturner07 View Post
    c. \int\frac{x^2}{\sqrt{1-x^2}} dx

    My Solution

    x = sin\theta, dx = cos\theta d\theta, \theta = \arcsin{x}

    = \int\frac{\sin^2(\theta)}{\sqrt{1 - \sin^2(\theta)}}cos\theta d\theta = \int\frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)}}co  s(\theta) d \theta = \int\frac{\sin^2(\theta)}{cos(\theta)}cos(\theta) d \theta = \int\sin^2(\theta) d\theta = \int\frac{1 {\color{red}+} cos(2\theta)}{2} d\theta = \frac{1}{2}\int 1 {\color{red}+} cos(2\theta) d\theta = \frac{1}{2}(\theta {\color{red}+} \frac{1}{2}\sin(2\theta)) = \frac{1}{2}\arcsin{x} {\color{red}+} \frac{1}{4}\sin({2{\color{red}x}}) + C
    See the red please.
    And really I like your way to post all work.
    I wonder if every asker can do that!
    Note: I did not check the other problems, It still maybe correct or wrong.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by mturner07 View Post
    d. \int\frac{x^3}{\sqrt{1 - x^2}} dx

    My Solution

    x = \sin(\theta), dx = \cos(\theta) d\theta, \theta = \arcsin{x}

    = \int\frac{\sin^3(\theta)}{\sqrt{1 - \sin^2(\theta)}}\cos(\theta) d\theta = \int\frac{\sin^3(\theta)}{\sqrt{\cos^2(\theta)}}\c  os(\theta) d\theta = \int\frac{\sin^3(\theta)}{cos(\theta)}\cos(\theta) d\theta = \int\sin^3(\theta) d\theta
    You just make it complicated from here.
    Notice that:
    \int\sin^3(\theta) d\theta=\int sin^2(\theta) sin(\theta) d\theta=\int (1-cos^2(\theta)) sin(\theta) d\theta
    Use u=cos(\theta).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by General View Post
    You just make it complicated from here.
    Notice that:
    \int\sin^3(\theta) d\theta=\int sin^2(\theta) sin(\theta) d\theta=\int (1-cos^2(\theta)) sin(\theta) d\theta
    Use u=cos(\theta).
    Note for (d) that you could also do the following:

    Let u=1-x^2\implies x^2=1-u. This implies that \,du=-2x\,dx

    The integral would then become -\tfrac{1}{2}\int\frac{1-u}{\sqrt{u}}\,du=\tfrac{1}{2}\int u^{1/2}-u^{-1/2}\,du
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    Posts
    28
    Thanks for the replies. Yeah it takes awhile to type up, but if people can't see my work they can't direct me to where I went wrong lol.

    Now for 2c. I get \frac{1}{2}\arcsin{x} - \frac{x}{4}\sqrt{1 - x^2} + C and for problem 2d. I get -\sqrt{1 - x^2} + \frac{(1 - x^2)\sqrt{1 - x^2}}{3} + C. Are these solutions, along with the other problems I posted correct? Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    28
    Anyone?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: May 14th 2010, 10:04 PM
  2. pappus
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 12th 2010, 10:02 AM
  3. reduction,centroid,pappus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2010, 06:42 AM
  4. Pappus theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 6th 2007, 12:01 AM
  5. Volume/Pappus' Theorem -- 30 mins remaning
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2006, 03:41 AM

Search Tags


/mathhelpforum @mathhelpforum